Calculus/Multivariable and differential calculus:Exercises

Polar Coordinates
Sketch the following polar curves without using a computer.

Sketch the following sets of points.

Calculus in Polar Coordinates
Find points where the following curves have vertical or horizontal tangents.

Sketch the region and find its area.

Cross Product
Find $$\mathbf u\times\mathbf v$$ and $$\mathbf v\times\mathbf u$$

Find the area of the parallelogram with sides $$\mathbf u$$ and $$\mathbf v$$.

Prove the following identities or show them false by giving a counterexample.

Length of Curves
Find the length of the following curves.

Limits And Continuity
Evaluate the following limits.

At what points is the function f continuous?

Use the two-path test to show that the following limits do not exist. (A path does not have to be a straight line.)

Partial Derivatives
Find the four second partial derivatives of the following functions.

Chain Rule
Find $$df/dt.$$

Find $$f_s,\ f_t.$$

Tangent Planes
Find an equation of a plane tangent to the given surface at the given point(s).

Maximum And Minimum Problems
Find critical points of the function f. When possible, determine whether each critical point corresponds to a local maximum, a local minimum, or a saddle point.

{{question-answer|question=260. $$f(x,y) = x^4 + 2y^2 - 4xy$$|answer={{noprint|1= The first order derivatives are: $$\frac{\partial f}{\partial x} = 4x^3 - 4y$$ and $$\frac{\partial f}{\partial y} = 4y - 4x$$. Finding the critical points is done by solving the equations $$\left\{\begin{array}{c} 4x^3 - 4y = 0 \\ 4y - 4x = 0 \end{array}\right.$$

$$4y - 4x = 0 \iff y = x$$. Substituting $$x$$ in place of $$y$$ in the first equation gives $$4x^3 - 4x = 0 \iff x(x^2-1) = 0 \iff x = -1, 0, +1$$. This gives the critical points $$(x,y) = (-1,-1), (0,0), (1,1)$$.

The second order derivatives are $$\frac{\partial^2 f}{\partial x^2} = 12x^2$$; $$\frac{\partial^2 f}{\partial y^2} = 4$$; and $$\frac{\partial^2 f}{\partial y \partial x} = -4$$. The discriminant is: $$\frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial y \partial x}\right)^2 = (12x^2)(4) - (-4)^2 = 48x^2 - 16$$.

For critical point $$(-1,-1)$$, the discriminant is $$32$$ and $$\frac{\partial^2 f}{\partial x^2} = 12$$ so $$(-1,-1)$$ is a local minimum.

For critical point $$(0,0)$$, the discriminant is $$-16$$ so $$(-1,-1)$$ is a saddle point.

For critical point $$(1,1)$$, the discriminant is $$32$$ and $$\frac{\partial^2 f}{\partial x^2} = 12$$ so $$(1,1)$$ is a local minimum.

Local minima at (1,1) and (−1,−1), saddle at (0,0)}}}}

{{question-answer|question=261. $$f(x,y) = \tan^{-1}(xy)$$|answer={{noprint|1= The first order derivatives are: $$\frac{\partial f}{\partial x} = \frac{1}{1+(xy)^2}y = \frac{y}{1+(xy)^2}$$ and $$\frac{\partial f}{\partial y} = \frac{1}{1+(xy)^2}x = \frac{x}{1+(xy)^2}$$. Finding the critical points is done by solving the equations $$\left\{\begin{array}{c} y/(1+(xy)^2) = 0 \\ x/(1+(xy)^2) = 0 \end{array}\right.$$

$$\frac{y}{1+(xy)^2} = 0 \iff y = 0$$ and $$\frac{x}{1+(xy)^2} = 0 \iff x = 0$$. So the only critical point is $$(x,y) = (0,0)$$.

The second order derivatives are $$\frac{\partial^2 f}{\partial x^2} = -\frac{y}{(1+(xy)^2)^2}(2xy^2) = \frac{-2xy^3}{(1+(xy)^2)^2}$$; $$\frac{\partial^2 f}{\partial y^2} = -\frac{x}{(1+(xy)^2)^2}(2x^2y) = \frac{-2x^3y}{(1+(xy)^2)^2}$$; and $$\frac{\partial^2 f}{\partial y \partial x} = \frac{(1)(1+(xy)^2) - (y)(2x^2y)}{(1+(xy)^2)^2} = \frac{1 - (xy)^2}{(1+(xy)^2)^2}$$. The discriminant is $$\frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial y \partial x}\right)^2 = \left(\frac{-2xy^3}{(1+(xy)^2)^2}\right)\left(\frac{-2x^3y}{(1+(xy)^2)^2}\right) - \left(\frac{1 - (xy)^2}{(1+(xy)^2)^2}\right)^2$$ $$ = \frac{4(xy)^4 - (1-(xy)^2)^2}{(1+(xy)^2)^4}$$ $$ = \frac{5(xy)^4 + 2(xy)^2 - 1}{(1+(xy)^2)^4}$$

For the critical point $$(0,0)$$, the discriminant is $$-1$$ so $$(0,0)$$ is a saddle point.

Saddle at (0,0)}}}}

{{question-answer|question=262. $$f(x,y) = 2xye^{-x^2-y^2}$$|answer={{noprint|1= The first order derivatives are: $$\frac{\partial f}{\partial x} = 2ye^{-x^2-y^2} + (2xy)e^{-x^2-y^2}(-2x) = (2y - 4x^2y)e^{-x^2-y^2}$$ and $$\frac{\partial f}{\partial y} = 2xe^{-x^2-y^2} + (2xy)e^{-x^2-y^2}(-2y) = (2x - 4xy^2)e^{-x^2-y^2}$$. Finding the critical points is done by solving the equations $$\left\{\begin{array}{c} (2y - 4x^2y)e^{-x^2-y^2} = 0 \\ (2x - 4xy^2)e^{-x^2-y^2} = 0 \end{array}\right.$$

$$\left\{\begin{array}{c} (2y - 4x^2y)e^{-x^2-y^2} = 0 \\ (2x - 4xy^2)e^{-x^2-y^2} = 0 \end{array}\right. \iff \left\{\begin{array}{c} 2y(1 - 2x^2) = 0 \\ 2x(1 - 2y^2) = 0 \end{array}\right.$$ $$\iff \left\{\begin{array}{c} y = 0 \;\text{or}\; x = -1/\sqrt{2} \;\text{or}\; x = 1/\sqrt{2} \\ x = 0 \;\text{or}\; y = -1/\sqrt{2} \;\text{or}\; y = 1/\sqrt{2} \end{array}\right.$$. The critical points are $$(x,y) = (0,0), (-1/\sqrt{2},-1/\sqrt{2}), (-1/\sqrt{2},1/\sqrt{2}), (1/\sqrt{2},-1/\sqrt{2}), (1/\sqrt{2},1/\sqrt{2})$$.

The second order derivatives are: $$\frac{\partial^2 f}{\partial x^2} = (-8xy)e^{-x^2-y^2} + (2y - 4x^2y)e^{-x^2-y^2}(-2x) = (8x^3y - 12xy)e^{-x^2-y^2}$$; $$\frac{\partial^2 f}{\partial y^2} = (-8xy)e^{-x^2-y^2} + (2x - 4xy^2)e^{-x^2-y^2}(-2y) = (8xy^3 - 12xy)e^{-x^2-y^2}$$; and $$\frac{\partial^2 f}{\partial y \partial x} = (2 - 4x^2)e^{-x^2-y^2} + (2y - 4x^2y)e^{-x^2-y^2}(-2y) = (8x^2y^2 - 4x^2 - 4y^2 + 2)e^{-x^2-y^2} = 2(2x^2-1)(2y^2-1)e^{-x^2-y^2}$$. The discriminant is $$\frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial y \partial x}\right)^2 = ((8x^3y - 12xy)e^{-x^2-y^2})((8xy^3 - 12xy)e^{-x^2-y^2}) - (2(2x^2-1)(2y^2-1)e^{-x^2-y^2})^2$$ $$ = (16x^2y^2(2x^2-3)(2y^2-3) - 4(2x^2-1)^2(2y^2-1)^2)e^{-2x^2-2y^2}$$.

For critical point $$(0,0)$$, the discriminant is $$-4(-1)(-1) = -4$$ so $$(0,0)$$ is a saddle point.

For critical points $$(-1/\sqrt{2},-1/\sqrt{2})$$ and $$(1/\sqrt{2},1/\sqrt{2})$$, the discriminant is $$4(-2)(-2)e^{-2} = 16e^{-2}$$ and $$\frac{\partial^2 f}{\partial x^2} = (2 - 6)e^{-1} = -4e^{-1}$$ so $$(-1/\sqrt{2},-1/\sqrt{2})$$ and $$(1/\sqrt{2},1/\sqrt{2})$$ are local maximums.

For critical points $$(-1/\sqrt{2},1/\sqrt{2})$$ and $$(1/\sqrt{2},-1/\sqrt{2})$$, the discriminant is $$4(-2)(-2)e^{-2} = 16e^{-2}$$ and $$\frac{\partial^2 f}{\partial x^2} = (-2 + 6)e^{-1} = 4e^{-1}$$ so $$(-1/\sqrt{2},1/\sqrt{2})$$ and $$(1/\sqrt{2},-1/\sqrt{2})$$ are local minimums.

Saddle at (0,0), local maxima at $$(\pm1/\sqrt2,\pm1/\sqrt2),$$ local minima at $$(\pm1/\sqrt2,\mp1/\sqrt2)$$}}}}

Find absolute maximum and minimum values of the function f on the set R. {{hidden|How to find candidate points for the absolute minimum and maximum of a function with a constrained domain.| Let $$f : \R^2 \to \R$$ denote the function for which the absolute minimum and maximum is sought. Let the domain be constrained to all points $$(x,y)$$ where $$g(x,y) = 0$$ where $$g(x,y)$$ is an appropriate function over $$\R^2$$.

Start with a candidate point $$(x,y) = (x_0,y_0)$$ where $$g(x_0,y_0) = 0$$, and envision that the $$x$$ and $$y$$ coordinates are changing at rates of $$m$$ and $$n$$ respectively: $$\frac{dx}{dt} = m$$ and $$\frac{dy}{dt} = n$$. The rate of change in $$f$$ is $$\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}$$ $$ = \frac{\partial f}{\partial x}m + \frac{\partial f}{\partial y}n$$. Since it is required that $$g(x,y) = 0$$, it must be the case that $$\frac{dg}{dt} = 0 \iff \frac{\partial g}{\partial x}\frac{dx}{dt} + \frac{\partial g}{\partial y}\frac{dy}{dt} = 0$$ $$ \iff \frac{\partial g}{\partial x}m + \frac{\partial g}{\partial y}n = 0$$.

$$(x,y) = (x_0,y_0)$$ is a local minimum or maximum only if $$\frac{df}{dt} = \frac{\partial f}{\partial x}m + \frac{\partial f}{\partial y}n = 0$$ for all $$(m,n)$$ where $$\frac{\partial g}{\partial x}m + \frac{\partial g}{\partial y}n = 0$$. This occurs iff the gradient $$\left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle$$ is parallel to the gradient $$\left\langle \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right\rangle$$. This condition can be quantified by $$\left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle = \lambda\left\langle \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right\rangle$$ where factor $$\lambda$$ is a "Lagrange multiplier".

Points $$(x,y)$$ where $$g(x,y) = 0$$ and $$\left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle = \lambda\left\langle \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right\rangle$$ for some $$\lambda$$ are candidate points for the absolute minimum or maximum. If the domain has any corners, then these corners are also candidate points. }}

{{question-answer|question=263. $$f(x,y) = x^2+y^2-2y+1,\ R=\{(x,y)\mid x^2+y^2\leq 4\}$$|answer={{noprint|1= Candidate points will be derived from two sources: Critical points of the function $$f(x,y) = x^2+y^2-2y+1$$ assuming an unconstrained domain, and candidate points assuming the restriction $$g(x,y) = x^2+y^2-4 = 0$$.

The first order derivatives of $$f(x,y)$$ are $$\frac{\partial f}{\partial x} = 2x$$ and $$\frac{\partial f}{\partial y} = 2y-2$$, so the only critical point where $$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0$$ occurs at $$(x,y) = (0,1)$$. This critical point lies in $$R$$ so it remains a valid candidate. $$f(0,1) = 0$$.

The first order derivatives of $$g(x,y)$$ are $$\frac{\partial g}{\partial x} = 2x$$ and $$\frac{\partial g}{\partial y} = 2y$$. Candidate points assuming the restriction $$g(x,y) = 0$$ must satisfy $$\left\{\begin{array}{c} x^2+y^2-4 = 0 \\ 2x = \lambda(2x) \\ 2y-2 = \lambda(2y) \end{array}\right.$$ for some $$\lambda$$. These equations are equivalent to $$\left\{\begin{array}{c} x^2+y^2-4 = 0 \\ x = 0 \;\text{or}\; \lambda = 1 \\ \lambda = (y-1)/y \end{array}\right.$$. If $$x = 0$$, then the only restriction left on $$y$$ is $$y^2-4 = 0$$ and $$y \neq 0$$. This gives two candidate points $$(0,-2), (0,2)$$. If $$\lambda = 1$$, then $$1 = (y-1)/y \iff y = y-1 \iff 0 = -1$$ which is never true. Hence the only valid candidate points derived by restricting the domain to $$g(x,y) = x^2+y^2-4 = 0$$ are $$(0,-2), (0,2)$$. $$f(0,-2) = 9$$ and $$f(0,2) = 1$$.

In total, the candidates are $$f(0,1) = 0$$, $$f(0,-2) = 9$$, and $$f(0,2) = 1$$.

The absolute minimum of $$0$$ occurs at $$(0,1)$$, and the absolute maximum of $$9$$ occurs at $$(0,-2)$$.

Maximum of 9 at (0,−2) and minimum of 0 at (0,1)}}}}

{{question-answer|question=264. $$f(x,y) = x^2+y^2-2x-2y,$$ R is a closed triangle with vertices (0,0), (2,0), and (0,2).|answer={{noprint|1= Triangle $$R$$ is defined by the constraints $$x \geq 0$$, $$y \geq 0$$, and $$x+y \leq 2$$.

Candidate points for the absolute minimum and maximum will be derived from 5 sources:
 * Critical points of the function $$f(x,y) = x^2+y^2-2x-2y$$ assuming an unconstrained domain.
 * Candidate points assuming the restriction $$g_1(x,y) = x = 0$$.
 * Candidate points assuming the restriction $$g_2(x,y) = y = 0$$.
 * Candidate points assuming the restriction $$g_3(x,y) = x+y-2 = 0$$.
 * The vertex points $$(0,0),(2,0),(0,2)$$.

The first order derivatives of $$f(x,y)$$ are $$\frac{\partial f}{\partial x} = 2x-2$$ and $$\frac{\partial f}{\partial y} = 2y-2$$, so the only critical point where $$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0$$ is $$(x,y) = (1,1)$$. The critical point $$(1,1)$$ lies in $$R$$ so it remains a valid candidate. $$f(1,1) = -2$$.

The first order derivatives of $$g_1(x,y)$$ are $$\frac{\partial g_1}{\partial x} = 1$$ and $$\frac{\partial g_1}{\partial y} = 0$$. Candidate points assuming the restriction $$g_1(x,y) = 0$$ must satisfy $$\left\{\begin{array}{c} x = 0 \\ 2x-2 = \lambda(1) \\ 2y-2 = \lambda(0) \end{array}\right.$$ for some $$\lambda$$. These equations are equivalent to $$\left\{\begin{array}{c} x = 0 \\ \lambda = -2 \\ y = 1 \end{array}\right.$$. This yields the candidate point $$(0,1)$$. Point $$(0,1)$$ lies in $$R$$ so it remains a valid candidate. $$f(0,1) = -1$$.

The first order derivatives of $$g_2(x,y)$$ are $$\frac{\partial g_2}{\partial x} = 0$$ and $$\frac{\partial g_2}{\partial y} = 1$$. Candidate points assuming the restriction $$g_2(x,y) = 0$$ must satisfy $$\left\{\begin{array}{c} y = 0 \\ 2x-2 = \lambda(0) \\ 2y-2 = \lambda(1) \end{array}\right.$$ for some $$\lambda$$. These equations are equivalent to $$\left\{\begin{array}{c} y = 0 \\ x = 1 \\ \lambda = -2 \end{array}\right.$$. This yields the candidate point $$(1,0)$$. Point $$(1,0)$$ lies in $$R$$ so it remains a valid candidate. $$f(1,0) = -1$$.

The first order derivatives of $$g_3(x,y)$$ are $$\frac{\partial g_3}{\partial x} = 1$$ and $$\frac{\partial g_3}{\partial y} = 1$$. Candidate points assuming the restriction $$g_3(x,y) = 0$$ must satisfy $$\left\{\begin{array}{c} x+y-2 = 0 \\ 2x-2 = \lambda(1) \\ 2y-2 = \lambda(1) \end{array}\right.$$ for some $$\lambda$$. The second equation yields $$\lambda = 2x-2$$. Substituting $$2x-2$$ in place of $$\lambda$$ in the third equation gives $$2y-2 = 2x-2 \iff y = x$$. Substituting $$x$$ in place of $$y$$ in the first equation gives $$x+x-2 = 0 \iff x=1$$, which then yields $$y=1$$ and $$\lambda=0$$. This yields the candidate point $$(1,1)$$. Point $$(1,1)$$ lies in $$R$$ so it remains a valid candidate. $$f(1,1) = -2$$.

Finally, we add the vertices $$(0,0),(2,0),(0,2)$$ to the lineup of candidate points.

Evaluating $$f(x,y) = x^2+y^2-2x-2y$$ at each candidate point gives $$f(1,1) = -2$$; $$f(0,1) = -1$$; $$f(1,0) = -1$$; $$f(0,0) = 0$$; $$f(2,0) = 0$$; and $$f(0,2) = 0$$. The absolute minimum of $$-2$$ occurs at $$(1,1)$$, while the absolute maximum of $$0$$ occurs at all of $$(0,0), (2,0), (0,2)$$.

Maximum of 0 at (2,0), (0,2), and (0,0); minimum of −2 at (1,1)}}}}

{{hidden|Finding the locations and shortest distances between two surfaces.| Consider two surfaces $$\sigma_1$$ and $$\sigma_2$$ in 3D space defined by the equations $$f_1(x,y,z) = 0$$ and $$f_2(x,y,z) = 0$$ respectively. Given a point $$(x_1,y_1,z_1)$$ from $$\sigma_1$$ and a point $$(x_2,y_2,z_2)$$ from $$\sigma_2$$, if $$(x_1,y_1,z_1)$$ and $$(x_2,y_2,z_2)$$ are the points that minimize the distance between $$\sigma_1$$ and $$\sigma_2$$, then it must be the case that the displacement $$\langle x_2-x_1, y_2-y_1, z_2-z_1 \rangle$$ is perpendicular to both surfaces. The gradient vector $$\left\langle \frac{\partial f_1}{\partial x}, \frac{\partial f_1}{\partial y}, \frac{\partial f_1}{\partial z} \right\rangle$$ is orthogonal to $$\sigma_1$$, and the gradient vector $$\left\langle \frac{\partial f_2}{\partial x}, \frac{\partial f_2}{\partial y}, \frac{\partial f_2}{\partial z} \right\rangle$$ is orthogonal to $$\sigma_2$$. The displacement vector must be parallel to both gradient vectors: $$\langle x_2-x_1, y_2-y_1, z_2-z_1 \rangle = \lambda_1\left\langle \frac{\partial f_1}{\partial x}, \frac{\partial f_1}{\partial y}, \frac{\partial f_1}{\partial z} \right\rangle = \lambda_2\left\langle \frac{\partial f_2}{\partial x}, \frac{\partial f_2}{\partial y}, \frac{\partial f_2}{\partial z} \right\rangle$$ for some $$\lambda_1$$ and $$\lambda_2$$.

Candidate points for the shortest distance between two surfaces must satisfy the following 8 equations: $$\left\{\begin{array}{c} f_1(x_1,y_1,z_1) = 0 \\ f_2(x_2,y_2,z_2) = 0 \\ x_2-x_1 = \lambda_1\frac{\partial f_1}{\partial x} \;\text{and}\; y_2-y_1 = \lambda_1\frac{\partial f_1}{\partial y} \;\text{and}\; z_2-z_1 = \lambda_1\frac{\partial f_1}{\partial z} \\ x_2-x_1 = \lambda_2\frac{\partial f_2}{\partial x} \;\text{and}\; y_2-y_1 = \lambda_2\frac{\partial f_2}{\partial y} \;\text{and}\; z_2-z_1 = \lambda_2\frac{\partial f_2}{\partial z} \end{array}\right.$$ for some $$\lambda_1$$ and $$\lambda_2$$. }}

{{question-answer|question=265. Find the point on the plane x−y+z=2 closest to the point (1,1,1).|answer={{noprint|1= The plane is defined by the equation $$f(x,y,z) = x-y+z-2 = 0$$, and a normal vector to the plane is $$\left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle = \langle 1, -1, 1 \rangle$$. The closest point $$(x,y,z)$$ on the plane to the point $$(1,1,1)$$ is the point where the displacement $$\langle x-1, y-1, z-1 \rangle$$ is parallel to $$\langle 1, -1, 1 \rangle$$. The following equations must be satisfied: $$\left\{\begin{array}{c} x-y+z-2 = 0 \\ x-1 = \lambda(1) \;\text{and}\; y-1 = \lambda(-1) \;\text{and}\; z-1 = \lambda(1) \end{array}\right.$$.

The second equation gives $$\lambda = x-1$$, and replacing $$\lambda$$ with $$x-1$$ in the third and forth equations gives $$y-1 = 1-x \iff y=2-x$$ and $$z-1=x-1 \iff z=x$$ respectively. In the first equation, replacing $$y$$ with $$2-x$$ and $$z$$ with $$x$$ gives $$x-(2-x)+x-2 = 0 \iff 3x - 4 = 0 \iff x = 4/3$$. This gives in turn $$y=2/3$$ and $$z=4/3$$.

The only candidate point for the closest distance is $$(4/3,2/3,4/3)$$, so therefore the point on the plane $$x-y+z=2$$ that is closest to the point $$(1,1,1)$$ is $$(4/3,2/3,4/3)$$.}}}}

{{question-answer|question=266. Find the point on the surface $$z = x^2+y^2+10$$ closest to the plane $$x+2y-z=0.$$|answer={{noprint|1= The surface is defined by the equation $$f_1(x,y,z) = x^2+y^2-z+10 = 0$$, and the plane is defined by the equation $$f_2(x,y,z) = x+2y-z = 0$$. Given a point $$(x_1,y_1,z_1)$$ from the surface and a point $$(x_2,y_2,z_2)$$ from the plane, these two points are closest to each other only if (but not always if) the displacement vector $$\langle x_2-x_1, y_2-y_1, z_2-z_1 \rangle$$ is a parallel to the surface normal vector $$\left\langle \frac{\partial f_1}{\partial x}, \frac{\partial f_1}{\partial y}, \frac{\partial f_1}{\partial z} \right\rangle = \langle 2x_1, 2y_1, -1 \rangle$$ and the plane normal vector $$\left\langle \frac{\partial f_2}{\partial x}, \frac{\partial f_2}{\partial y}, \frac{\partial f_2}{\partial z} \right\rangle = \langle 1, 2, -1 \rangle$$. There must exist factors $$\lambda_1$$ and $$\lambda_2$$ such that the following 8 equations hold: $$\left\{\begin{array}{c} x_1^2+y_1^2-z_1+10 = 0 \\ x_2+2y_2-z_2 = 0 \\ x_2-x_1 = \lambda_1(2x_1) \;\text{and}\; y_2-y_1 = \lambda_1(2y_1) \;\text{and}\; z_2-z_1 = \lambda_1(-1) \\ x_2-x_1 = \lambda_2(1) \;\text{and}\; y_2-y_1 = \lambda_2(2) \;\text{and}\; z_2-z_1 = \lambda_2(-1) \end{array}\right.$$

The fifth equation is equivalent to $$\lambda_1 = -(z_2-z_1)$$, and the eighth equation is equivalent to $$\lambda_2 = -(z_2-z_1)$$. Eliminating via substitution $$\lambda_1$$ and $$\lambda_2$$ in all of the other equations gives: $$\left\{\begin{array}{c} x_1^2+y_1^2-z_1+10 = 0 \\ x_2+2y_2-z_2 = 0 \\ x_2-x_1 = -2x_1(z_2-z_1) \;\text{and}\; y_2-y_1 = -2y_1(z_2-z_1) \\ x_2-x_1 = -(z_2-z_1) \;\text{and}\; y_2-y_1 = -2(z_2-z_1) \end{array}\right.$$

If $$z_2-z_1 = 0$$, then $$x_2-x_1 = 0$$ and $$y_2-y_1 = 0$$, and then $$(x_1,y_1,z_1)$$ and $$(x_2,y_2,z_2)$$ are the same point which corresponds to an intersection between the surface and the plane. While it will not be shown here, it is relatively simple to show that the surface and plane fail to intersect. Excluding the possibility that $$z_2-z_1 = 0$$, the equations $$x_2-x_1 = -2x_1(z_2-z_1)$$ and $$x_2-x_1 = -(z_2-z_1)$$ together imply that $$-2x_1 = -1 \iff x_1 = 1/2$$; and the equations $$y_2-y_1 = -2y_1(z_2-z_1)$$ and $$y_2-y_1 = -2(z_2-z_1)$$ together imply that $$-2y_1 = -2 \iff y_1 = 1$$. The values $$x_1 = 1/2$$ and $$y_1 = 1$$ give $$z_1 = x_1^2+y_1^2+10 = (1/2)^2 + 1^2 + 10 = 11.25$$. Hence the point on the surface $$z = x^2+y^2+10$$ that is closest to the plane $$x+2y-z=0$$ is $$(x_1,y_1,z_1) = (0.5,1,11.25)$$.

From $$x_2-x_1 = -(z_2-z_1)$$ and $$y_2-y_1 = -2(z_2-z_1)$$, it follows that $$x_2-0.5 = -(z_2-11.25) \iff x_2 = 11.75 - z_2$$ and $$y_2-1 = -2(z_2-11.25) \iff y_2 = 23.5 - 2z_2$$. In the equation $$x_2+2y_2-z_2 = 0$$, eliminating $$x_2$$ and $$y_2$$ via substitution gives $$(11.75-z_2) + 2(23.5-2z_2) - z_2 = 0 \iff 6z_2 = 58.75 \iff z_2 = 9.791\overline{6}$$. Hence $$x_2 = 1.958\overline{3}$$ and $$y_2 = 3.91\overline{6}$$. The corresponding closest point on the plane is $$(x_2,y_2,z_2) = (1.958\overline{3}, 3.91\overline{6}, 9.791\overline{6})$$.

The closest point on the surface is $$(0.5, 1, 11.25)$$.}}}}

Double Integrals over Rectangular Regions
Evaluate the given integral over the region R.

Evaluate the given iterated integrals.

Double Integrals over General Regions
Evaluate the following integrals.

Use double integrals to compute the volume of the given region.

Triple Integrals
In the following exercises, sketching the region of integration may be helpful.

Vector Fields
One can sketch two-dimensional vector fields by plotting vector values, flow curves, and/or equipotential curves.

Conservative Vector Fields
Determine if the following vector fields are conservative on $$\mathbb R^2.$$

Determine if the following vector fields are conservative on their respective domains in $$\mathbb R^3.$$ When possible, find the potential function.

Divergence Theorem
Compute the net outward flux of the given field across the given surface.